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2 years ago

For the common-emitter, common-base and emitter-follower amplifier designs,what is the primary benefit of each amplifier?

Physics

1 answer:

kati45 [8]2 years ago

6 0

Answer:

Explanation:

A common emitter amplifier works by inverting. It does have low input impedance. Despite its low input impedance, it poses an otherwise high output impedance.

The common base circuit performs optimally when it acts as a current buffer. It has the ability to take an input current at a low input impedance, and transmit almost the same current to an impedance with a higher output

The primary benefit of emitter follower amplifier is that the transistor is able to provide current and power gain. Although this transistor takes in little current from the input. It still provides an impedance with a low output to a circuit by exercising the output of the follower. This then translates to that the output under load not dropping.

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A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

Nimfa-mama [501]

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

here we know that natural frequency of the oscillation is given as

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

$\omega = 2\pi(10.5) = 65.97 rad/s$

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$

6 0

3 years ago

A certain circuit is composed of two series resistors. The total resistance is 10 Ohms. One of the resistors is 4 Ohms. The othe

Mariulka [41]

<h3>Given :- </h3>

A certain circuit is composed of two series resistors
The total resistance is 10 ohms
One of the resistor is 4 ohms

We have to find the value of other resistor?

<h3>Let's Begin :-</h3>

We know that,

In series combination,

When a number of resistances are connected in series, the equivalent I.e resultant resistance is equal to the sum of the individual resistances and is greater than any individual resistance

That is,

Rn in series = R1 + R2 + R3.....So on

Therefore,

According to the question,

We have,

R1 + R2 = 10 Ω

4 + R2 = 10Ω

R2 = 10 - 4

R2 = 6Ω

Hence, The value of R2 resistor in series is 6Ω